A student considers how many people enter a particular bank during the lunch hour and how many exit. She decides to run a simulation. She assigns the value 0 to represent a person entering the bank and 1 to represent a person exiting the bank. Based on a simulation for the first 20 people she observes, estimate the probability she will see a person entering the bank. \( 0110 \begin{array}{lllllllllllllll}0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0\end{array} 1 \) (1 point) 0.400 0.250

To estimate the probability of a person entering the bank based on the simulation, we need to analyze the sequence of values provided. The values represent: - 0: a person entering the bank - 1: a person exiting the bank The sequence given is: \( 0110 \, 0 \, 0 \, 1 \, 1 \, 1 \, 0 \, 1 \, 0 \, 0 \, 0 \, 1 \, 0 \, 1 \, 0 \, 0 \, 1 \) ### Step 1: Count the total number of entries and exits 1. Count the number of 0s (entries): - From the sequence, we have: \( 0, 0, 0, 0, 0, 0, 0 \) (7 entries) 2. Count the number of 1s (exits): - From the sequence, we have: \( 1, 1, 1, 1, 1, 1, 1, 1, 1 \) (13 exits) ### Step 2: Calculate the total number of people observed The total number of people observed is the sum of entries and exits: \[ \text{Total} = \text{Number of 0s} + \text{Number of 1s} = 7 + 13 = 20 \] ### Step 3: Calculate the probability of a person entering the bank The probability \( P \) of a person entering the bank is given by the ratio of the number of entries to the total number of people observed: \[ P(\text{entering}) = \frac{\text{Number of 0s}}{\text{Total}} = \frac{7}{20} \] ### Step 4: Convert to decimal Calculating the decimal value: \[ P(\text{entering}) = \frac{7}{20} = 0.35 \] ### Conclusion The estimated probability of seeing a person entering the bank is \( 0.35 \). However, since the options provided are \( 0.400 \) and \( 0.250 \), neither of these options matches our calculated probability. Thus, the answer is not among the provided options.